Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $n = \dfrac{-4a^2 - 28a - 24}{2a^2 - 4a - 96} \times \dfrac{-2a + 16}{4a - 24} $
Explanation: First factor out any common factors. $n = \dfrac{-4(a^2 + 7a + 6)}{2(a^2 - 2a - 48)} \times \dfrac{-2(a - 8)}{4(a - 6)} $ Then factor the quadratic expressions. $n = \dfrac {-4(a + 6)(a + 1)} {2(a + 6)(a - 8)} \times \dfrac {-2(a - 8)} {4(a - 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { -4(a + 6)(a + 1) \times -2(a - 8)} { 2(a + 6)(a - 8) \times 4(a - 6)} $ $n = \dfrac {8(a + 6)(a + 1)(a - 8)} {8(a + 6)(a - 8)(a - 6)} $ Notice that $(a + 6)$ and $(a - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {8\cancel{(a + 6)}(a + 1)(a - 8)} {8\cancel{(a + 6)}(a - 8)(a - 6)} $ We are dividing by $a + 6$ , so $a + 6 \neq 0$ Therefore, $a \neq -6$ $n = \dfrac {8\cancel{(a + 6)}(a + 1)\cancel{(a - 8)}} {8\cancel{(a + 6)}\cancel{(a - 8)}(a - 6)} $ We are dividing by $a - 8$ , so $a - 8 \neq 0$ Therefore, $a \neq 8$ $n = \dfrac {8(a + 1)} {8(a - 6)} $ $ n = \dfrac{a + 1}{a - 6}; a \neq -6; a \neq 8 $